A110021 -id:A110021 - cp's OEIS frontend

A071866 Number of elements in the continued fraction for prime(n+1)/prime(n).

2, 3, 3, 4, 3, 3, 3, 4, 4, 3, 3, 3, 3, 4, 4, 4, 3, 3, 4, 3, 3, 4, 4, 3, 3, 3, 4, 3, 3, 3, 4, 4, 3, 4, 3, 3, 3, 4, 4, 4, 3, 3, 3, 3, 3, 6, 6, 4, 3, 3, 4, 3, 3, 4, 4, 4, 3, 3, 3, 3, 4, 4, 4, 3, 3, 6, 3, 5, 3, 3, 4, 4, 3, 3, 4, 4, 6, 3, 3, 4, 3, 3, 3, 3, 4, 4, 3, 3, 3, 4, 4, 4, 4, 5, 4, 4, 5, 3, 3, 3, 5, 4, 4, 3, 3

Offset: 1

Benoit Cloitre, Jun 09 2002

			prime(5)/prime(4) = 11/7, 11/7 continued fraction is [1, 1, 1, 3] which contains 4 elements, hence a(4)=4.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A110021, A109374, A112323, A112324, A112768.

seq(nops(convert(ithprime(n+1)/ithprime(n),confrac)),n=1..200); # Robert Israel, May 29 2018

Table[Length[ContinuedFraction[Prime[n + 1]/Prime[n]]], {n, 105}] (* Ray Chandler, Sep 18 2005 *)

a(n) = length(contfrac(prime(n+1)/prime(n)));

More terms from Hans Havermann, Jul 06 2002

A109374 Irregular table read by rows: Row n is the terms of the continued fraction for prime(n+1)/prime(n).

Original entry on oeis.org

1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 3, 1, 5, 2, 1, 3, 4, 1, 8, 2, 1, 4, 1, 3, 1, 3, 1, 5, 1, 14, 2, 1, 5, 6, 1, 9, 4, 1, 20, 2, 1, 10, 1, 3, 1, 7, 1, 5, 1, 8, 1, 5, 1, 29, 2, 1, 10, 6, 1, 16, 1, 3, 1, 35, 2, 1, 12, 6, 1, 19, 1, 3, 1, 13, 1, 5, 1, 11, 8, 1, 24, 4, 1, 50, 2, 1, 25, 1, 3, 1, 53, 2, 1, 27, 4, 1

Offset: 1

Leroy Quet, Aug 24 2005

Sequence A071866 gives the number of terms in the n-th continued fraction.

If n is in A029707, row n is [1, (prime(n)-1)/2, 2]. - Robert Israel, May 29 2018

			Prime(4)/prime(3) = 7/5 = 1+ 1/(2+1/2), so the terms associated with the 3rd continued fraction are 1, 2, 2.

Robert Israel, Table of n, a(n) for n = 1..10003 (rows 1 to 2533, flattened)

Cf. A029707, A071866, A110021, A112323, A112324, A112768.

seq(op(convert(ithprime(n+1)/ithprime(n), confrac)),n=1..100); # Robert Israel, May 29 2018

Flatten[Table[ContinuedFraction[Prime[n + 1]/Prime[n]], {n, 30}]] (* Ray Chandler, Aug 25 2005 *)

Extended by Ray Chandler and Robert G. Wilson v, Aug 25 2005
Edited by Charles R Greathouse IV, Apr 23 2010
Definition corrected by Leroy Quet, May 10 2010

A112323 a(n) = numerator of sum of reciprocals of the terms of the continued fraction for prime(n+1)/prime(n).

Original entry on oeis.org

3, 5, 2, 10, 17, 19, 13, 31, 38, 11, 41, 49, 31, 73, 82, 93, 89, 19, 115, 107, 5, 136, 148, 107, 31, 38, 178, 161, 139, 67, 220, 236, 103, 256, 56, 181, 47, 283, 302, 313, 269, 52, 287, 61, 74, 65, 69, 388, 341, 289, 423, 359, 137, 456, 467, 478, 101, 107, 349, 211

Offset: 1

Leroy Quet, Sep 03 2005

			prime(6)/prime(5) = 13/11 = 1 + 1/(5 + 1/2).
So a(5) is 17, the numerator of 17/10 = 1 + 1/5 + 1/2.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A071866, A110021, A109374, A112324, A112768.

f[n_] := Plus @@ (1/# &) /@ ContinuedFraction[Prime[n + 1]/Prime[n]]; Table[Numerator[f[n]], {n, 60}] (* Ray Chandler, Sep 07 2005 *)

Extended by Ray Chandler, Sep 07 2005

A112768 Sum of terms of continued fraction for prime(n+1)/prime(n).

Original entry on oeis.org

3, 4, 5, 6, 8, 8, 11, 9, 10, 17, 12, 14, 23, 15, 14, 15, 32, 17, 21, 38, 19, 24, 20, 20, 29, 53, 30, 56, 32, 23, 36, 28, 71, 24, 77, 32, 33, 45, 34, 35, 92, 29, 98, 53, 101, 23, 24, 60, 116, 62, 45, 122, 35, 48, 49, 50, 137, 52, 74, 143, 35, 35, 81, 158, 83, 30, 62, 40, 176

Offset: 1

Ray Chandler, Sep 18 2005

			Prime(6)/prime(5) = 13/11 = 1 + 1/(5 + 1/2), so a(5) = 1+5+2 = 8.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A071866, A110021, A109374, A112323, A112324.

seq(convert(numtheory:-cfrac(ithprime(n+1)/ithprime(n), quotients),`+`),n=1..100); # Robert Israel, Jul 11 2017

Table[Plus @@ ContinuedFraction[Prime[n + 1]/Prime[n]], {n, 70}]

a(n) = vecsum(contfrac(prime(n+1)/prime(n))); \\ Michel Marcus, Jul 12 2017

A112324 a(n) = denominator of sum of reciprocals of the terms of the continued fraction for prime(n+1)/prime(n).

Original entry on oeis.org

2, 2, 1, 3, 10, 12, 8, 12, 15, 7, 30, 36, 20, 30, 35, 40, 58, 15, 48, 70, 4, 57, 65, 88, 24, 25, 75, 106, 108, 56, 93, 105, 68, 117, 37, 150, 39, 120, 135, 140, 178, 45, 190, 48, 49, 16, 17, 165, 226, 228, 190, 238, 120, 205, 210, 215, 67, 90, 276, 140, 84, 260, 228, 310

Offset: 1

Leroy Quet, Sep 03 2005

			prime(6)/prime(5) = 13/11 = 1 + 1/(5 + 1/2).
So a(5) is 10, the denominator of 17/10 = 1 + 1/5 + 1/2.

Cf. A071866, A110021, A109374, A112323, A112768.

f[n_] := Plus @@ (1/# &) /@ ContinuedFraction[Prime[n + 1]/Prime[n]]; Table[Denominator[f[n]], {n, 64}] (* Ray Chandler, Sep 07 2005 *)

Extended by Ray Chandler, Sep 07 2005

cp's OEIS Frontend

A071866 Number of elements in the continued fraction for prime(n+1)/prime(n).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Extensions

A109374 Irregular table read by rows: Row n is the terms of the continued fraction for prime(n+1)/prime(n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Extensions

A112323 a(n) = numerator of sum of reciprocals of the terms of the continued fraction for prime(n+1)/prime(n).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Extensions

A112768 Sum of terms of continued fraction for prime(n+1)/prime(n).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

A112324 a(n) = denominator of sum of reciprocals of the terms of the continued fraction for prime(n+1)/prime(n).

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Extensions